I have a question for a coefficient test of poisson regression. That test uses a Wald statistic defined as;
$$ W = \frac{\hat{\beta}-\beta_0}{\hat{se}(\hat{\beta)}} $$
where $\hat{\beta}$, $\hat{se}(\hat{\beta)}$ is MLE, standard error. This answer says the variance of the residuals is related to mean in poisson regression (also logistic one). That's why this Wald statistic will be tested by z-test.
However, I can't understand that's point. Why do we use z-test? I understand poisson distribution's mean and variance are related respectively ($Y\sim Po(\lambda)$, then $E(Y)=Var(Y)=\lambda$). But this relationship is for $Y$, NOT $\beta$, right? If that's the case, this relationship($E(Y)=Var(Y)=\lambda$) is unrelated to $W$?
Does $\beta$ follow Poisson distribution? Is $\beta$ a random variable? Do we assume that $\beta$'s MLE's variance is related to that's mean? Oh, I'm confused...
Also, considering $\hat{\beta}$'s asymptotic theory, the below formula is satisfied;
$$ \sqrt{n}(\hat{\beta} - \beta_0) \rightarrow \mathcal{N}(0,\frac{1}{I(\beta))}) $$
where $I(\beta))$ is fisher information. Therefore, I think the Wald statistic's $\hat{se}(\hat{\beta)}$ may represent $\frac{I(\beta)}{\sqrt{n}}$. Is it correct?
Summary:
- Why do we use z-test for coefficient test of poisson (logistic) regression?
- Why is a coefficient's MLE($\hat{\beta}$)'s variance related to that's mean?
- Does Wald statistic's standard error $\hat{se}(\hat{\beta)}$ represent $\frac{I(\beta)}{\sqrt{n}}$?
